The proof of Proposition REF  given in [1]}
consists to interpret members of the age by words over a finite alphabet and apply Higman's Theorem on words. Here, using the same alphabet, we interpret the countable structures whose ages are included in \(\mathfrak {C}\)  by countable chains labelled by this alphabet. Since the alphabet is b.q.o., Laver's Theorem implies that this collection of labelled chains is b.q.o. Now, if we label these labelled chains by a b.q.o., an other application of Laver's Theorem yields that the resulting class is b.q.o. Here are the details.
We asserts that \(\tilde{\mathfrak {C}}_{\le \omega }\)  is the image of the product \(\mathfrak {P}\)  of a finite poset and the class \(\mathcal {S}_{\le \omega }\)  labelled by a finite set \(A\)  by a map \(\varphi \)  preserving the order. Any finite poset being b.q.o., \(\mathcal {S}_{\le \omega }.A\)  is b.q.o. via Laver's Theorem, the product \(\mathfrak {P}\)  is b.q.o. and its image too, hence \(\tilde{\mathfrak {C}}_{\le \omega }\)  is b.q.o. If we label members of \(\tilde{\mathfrak {C}}_{\le \omega }\)  by a b.q.o., say \(Q\) , this amounts to label the members of \(\mathfrak {P}\)  by \(Q\) . An other application of Laver's theorem yields that \(\mathfrak {P}.Q\)  is b.q.o., hence
\(\tilde{\mathfrak {C}}_{\le \omega }. Q\)  is b.q.o. as required. To prove our assertion, let \(R:=: (V, (\rho _i)_{i\in I})\)  be an almost multichainable structure as defined above
such that \(Age(R)= \mathfrak {C}\) . The set \(V\)  is the disjoint union of a finite set \(F\)  and a set of the form \(L\times K\)  where \(K\)  is a finite set and there is a linear order on \(L\)  satisfying condition \(\bullet \)  above. We code \(R\)  by an almost multichain \(S:= (V, \le , \rho _L, \rho _K, a_1, \dots a_{\ell })\)  where the elements \(a_1, \dots , a_k\)  are constants defining \(F\)  that is \(\lbrace a_1, \dots , a_k\rbrace =F\) , the relations \(\rho _L\)  and \(\rho _K\)  are equivalence relations on \(V\)  such that all the elements of \(F\)  are inequivalent and inequivalent to those of \(L\times K\) , furthermore two elements \((x,y)\)  and \((x^{\prime },y^{\prime })\)  of \(L\times K\)  are equivalent with respect to \(\rho _K\)  if \(y=y^{\prime }\) , respectively \(\rho _L\)  if \(x=x^{\prime }\) ; the relation \(\le \)  is an order on \(V\)  for which all the elements of \(F\)  are minimal and below all elements of \(L\times K\) , all blocks of \(\rho _L\)  are antichains for the ordering and the family of these blocks is totally ordered by \(\le \) . Let \(\mathfrak {D}:= Age( S)\)  and \(\tilde{\mathfrak {D}}_{\le \omega } \)  be the collection of countable structures whose age is included in \(\mathfrak {D}\) . We may apply Compactness Theorem of First Order Logic to this elaborate construction. It yields that for every member \(R^{\prime }\)  of \(\mathfrak {C}= Age(R)\)  there is some \(S^{\prime }\)  on the same domain such that \(Age(S^{\prime }) \subseteq Age(S)= \mathfrak {D}\)  and every local isomorphism from \(S^{\prime }\)  to \(S\)  yields a local isomorphism from \(R^{\prime }\)  to \(R\)  (beware, \(S^{\prime }\)  is not necessarily almost multichainable).
Thus \(\tilde{\mathfrak {D}}_{\le \omega }\)  is transformed in \(\tilde{\mathfrak {C}}_{\le \omega }\)  in an order preserving way. Hence, if \(Q\)  is a b.q.o., in order to prove that \(\tilde{\mathfrak {C}}_{\le \omega }. Q\)  is b.q.o., it suffices to prove that \(\tilde{\mathfrak {D}}_{\le \omega }. Q\) . is b.q.o. For
this new step, let \(A\)  be the alphabet whose letters are the non-empty subsets of \(K\)  (that is
\(A := (K ) \setminus \lbrace \emptyset \rbrace \) . We order \(A\)  by inclusion and
extends this order to the collection \(\mathcal {S}_{\le \omega }. A\)  of countable chains labelled by the elements of \(A\) . Order \((F)\)  by inclusion. Then \(\tilde{\mathfrak {D}}_{\le \omega }\)  is the image of the direct product \(\mathfrak {P}:= (F) \times \mathcal {S}_{\le \omega }. A\)  (thus, via Laver's Theorem, \(\tilde{\mathfrak {D}}_{\le \omega }\)  is b.q.o. ). Then, replace \(A\)  by the set \(A^{\prime }\)  of \(f: X\rightarrow Q\)  such that \(X\in A\)  in this direct product. An other application of Laver's theorem yields that \(\tilde{\mathfrak {D}}_{\le \omega }. Q\)  is b.q.o.
